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To solve this problem, we need to find the rate of change of the height of the liquid with respect to time (\(\frac{dh}{dt}\)) at the moment when \( h = 3 \) inches. We are given that the radius \( r \) of the funnel at height \( h \) is given by the equation:

\[
r = \frac{1}{20}(3 + h^2)
\]

Additionally, we know that at \( h = 3 \) inches, the radius \( r \) is decreasing at a rate of \( \frac{dr}{dt} = -\frac{1}{5} \) inches per second. Our goal is to find \( \frac{dh}{dt} \).

### Step 1: Differentiate the equation for \( r \) with respect to time \( t \)

We start by differentiating the equation for \( r \) with respect to \( t \):

\[
r = \frac{1}{20}(3 + h^2)
\]

Taking the derivative with respect to \( t \) using the chain rule:

\[
\frac{dr}{dt} = \frac{1}{20} \cdot 2h \cdot \frac{dh}{dt}
\]

This simplifies to:

\[
\frac{dr}{dt} = \frac{h}{10

The inside of a funnel of height 10 inches has circular cross-sections. At height h, the radius of the funnel is given by r = 1/20(3 + h^2), where 0 ≤ h ≤ 10. The funnel contains liquid that is draining from the bottom. At the instant when the height of the liquid is h = 3 inches, the radius of the surface of the liquid is decreasing at a rate of 1/5 inch per second. What is the rate of change of the height of the liquid with respect to time?

Explore a related rates problem involving the rate of change of liquid height in a funnel. This problem utilizes differentiation and the chain rule to determine the rate at which the height of the liquid changes with time, given the rate of change of the radius. Ideal for students in Grades 11-12.

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